## 1.1 References

• Stichtenoth.

# 2 Lecture 1: Field Theory Preliminaries

The main theorems in this course, in order of importance:

• The Riemann-Roch Theorem
• The Riemann-Hurwitz Formula

## 2.1 Finite Generation of Fields

See Chapter 11 of Field Theory notes.

### 2.1.1 Notion 1

A field extension $$\ell/k$$ is finitely generated if there exists a finite set $$x_1, \cdots, x_n \in \ell$$ such that $$\ell = k(x_1, \cdots, x_n)$$ and $$\ell$$ is the smallest field extension of $$k$$.

Concretely, every element of $$\ell$$ is a quotient of the form $${p(x_1, \cdots, x_n) \over q(x_1, \cdots, x_n)}$$ with $$p, q\in k[x_1, \cdots, x_n]$$.

There are three different notions of finite generation for fields, the above is the weakest.

### 2.1.2 Notion 2

The second is being finitely generated as an algebra:

For $$R\subset S$$ finitely generated algebras, $$S$$ is finitely generated over $$R$$ if every element of $$S$$ is a polynomial in $$x_1, \cdots, x_n$$, with coefficients in $$R$$, i.e. $$S = R[x_1, \cdots, x_n]$$.

Note that this implies the previous definition, since anything that is a polynomial is also a quotient of polynomials.

### 2.1.3 Notion 3

The final notion: $$\ell/k$$ is finite (finite degree) if $$\ell$$ is finitely generated as a $$k{\hbox{-}}$$module, i.e. a finite-dimensional $$k{\hbox{-}}$$vector space.

A rational function field is $$k(t_1, \cdots, t_n) \mathrel{\vcenter{:}}= ff \qty{ k[t_1, \cdots, t_n]}$$.

Note that we can make a similar definition for infinitely many generators by taking a direct limit (here: union), and in fact every element will only involve finitely many generators.

1. Show $$k(t) / k$$ is finitely generated by notion (3) but not by (2).

2. Show that $$k[t]/k$$ is (2) but not (1).1

3. Show that it is not possible for a field extension to satisfy (2) but not (1).2

4. Show that if $$\ell/k$$ is finitely generated by (3) and algebraic, then it satisfies (1).

If $$L/K/F$$ are field extensions, then $$L/F$$ is finitely generated $$\iff$$ $$K/F$$ and $$L/K$$ are finitely generated.3

For $$\ell/k$$, a subset $$\left\{{x_i}\right\}\subset \ell$$ is algebraically independent over $$k$$ if no finite subset satisfies a nonzero polynomial with $$k$$ coefficients.

In this case, $$k[\left\{{x_i}\right\}] / k$$ is purely transcendental as a rational function field.

For $$\ell/k$$ a field extension,

1. There exists a subset $$\left\{{x_i}\right\}\subset \ell$$ algebraically independent over $$k$$ such that $$\ell/k(\left\{{x_i}\right\})$$ is algebraic.

2. If $$\left\{{y_t}\right\}$$ is another set of algebraically independent elements such that $$\ell/k(\left\{{y_t}\right\})$$ is algebraic, then $${\left\lvert {\left\{{x_i}\right\}} \right\rvert} = {\left\lvert {\left\{{y_t}\right\}} \right\rvert}$$.

Thus every field extension is algebraic over a purely transcendental extension. A subset as above is called a transcendence basis, and every 2 such bases have the same cardinality.

We have a notion of generation (similar to “spanning”), independence, and bases, so there are analogies to linear algebra (e.g. every vector space has a basis, any two have the same cardinality).4

The following notion will be analagous to that of dimension in linear algebra:

The transcendence degree of $$\ell/k$$ is the cardinality of any transcendence basis.

If $$L/K/F$$ are fields then $$\operatorname{trdeg}(L/F) = \operatorname{trdeg}(K/F) + \operatorname{trdeg}(L/K)$$.

Let $$K/k$$ be finitely degenerated, so $$K = k(x_1, \cdots, x_n)$$. Then $$\operatorname{trdeg}(K/k) \leq n$$, with equality iff $$K/k$$ is purely transcendental.

Suppose $$K$$ is monogenic, i.e. generated by one element. Then $$\operatorname{trdeg}(F(x)/F) = \indic{x/F\text{ is transcendental}}$$.

So the degree increases when a transcendental element is added, and doesn’t change when $$x$$ is algebraic.

By additivity in towers, we take $$k \hookrightarrow k(x_1) \hookrightarrow k(x_1, x_2) \hookrightarrow\cdots \hookrightarrow k(x_1, \cdots, x)n) = K$$ to obtain a chain of length $$n$$. The transcendence degree is thus the number of indices $$i$$ such that $$x_i$$ is transcendental over $$k(x_1, \cdots, x_{i-1})$$.5

For $$d\in{\mathbb{Z}}^{\geq 0}$$, an extension $$K/k$$ is a function field in $$d$$ variables (i.e. of dimension $$d$$) if $$K/k$$ is finitely generated of transcendence degree $$d$$.

The study of such fields is birational geometry over the ground field $$k$$. $$k={\mathbb{C}}$$ is of modern interest, things get more difficult in other fields. The case of $$d=1$$ is much easier: the function field will itself be the geometric object and everything will built from that. Our main tool will be valuation theory, where valuations will correspond to points on the curve.

## 2.2 Case Study: The Lüroth Problem.

For which fields $$k$$ and $$d\in {\mathbb{Z}}^{\geq 0}$$ is it true that if $$k \subset \ell \subset k(t_1, \cdots, t_d)$$ with $$k(t_1 ,\cdots, t_d)/\ell$$ finite then $$\ell$$ is purely transcendental?

It’s complicated, and depends on $$d$$ and $$k$$. We have the following partial results.

True for $$d=1$$: For any $$k\subset \ell \subset k(t)$$, $$\ell = k(x)$$.

Also true for $$d=2, k={\mathbb{C}}$$.

No if $$d= 2$$, $$k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$, and $$k$$ is positive characteristic. Also no if $$d=2, k\neq \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ in characteristic zero.

No if $$d\geq 3$$ and $$k= {\mathbb{C}}$$.

Note that unirational need not imply rational for varieties.

Let $$k$$ be a field, $$G$$ a finite group with $$G\hookrightarrow S_n$$ the Cayley embedding. Then $$S_n$$ acts by permutation of variables on $$k(t_1, \cdots, t_n)$$, thus so does $$G$$. Set $$\ell \mathrel{\vcenter{:}}= k(t_1, \cdots, t_n)^G$$ the fixed field, then by Artin’s observation in Galois theory: if you have a finite field acting effectively by automorphisms on a field then taking the fixed field yields a galois extension with automorphism group $$G$$.

So $$\operatorname{Aut}(k(t_1, \cdots, t_n)/ \ell) = G$$.A

1. Suppose $$k={\mathbb{Q}}$$, and show that an affirmative answer to the Lüroth problem implies an affirmative answer to the inverse galois problem for $${\mathbb{Q}}$$.

Hint: works for any field for which Hilbert’s Irreducibility Theorem holds.

1. $$\ell /{\mathbb{Q}}$$ need not be a rational function field, explore the literature on this: first example due to Swan with $${\left\lvert {G} \right\rvert} = 47$$.

2. Can still give many positive examples using the Shepherd-Todd Theorem.

## 2.3 Integrals Closures and Constant Fields

For $$K/k$$ a field extension, set $$\kappa(K)$$ to be the algebraic closure of $$k$$ in $$K$$, i.e. special case of integral closure. If $$K/k$$ is finitely generated, then $$\kappa(K)/k$$ is finite degree.

Here $$\kappa(K)$$ is called the field of constants, and $$K$$ is also a function field over $$\kappa(K)$$.

In practice, we don’t want $$\kappa(K)$$ to be a proper extension of $$k$$. If this isn’t the case, we replace considering $$K/k$$ by $$K/\kappa(K)$$. If $$K/k$$ is finitely generated, then Where we use the fact that from above, $$\kappa(K)/k$$ is finitely generated and algebraic and thus finite, and by a previous theorem, if $$K/k$$ is transcendental then $$K/\kappa(K)$$ is as well, and thus finitely generated. Thus if you have a function field over $$k$$, you can replace $$k$$ by $$\kappa(K)$$ and regard $$K$$ as a function field over $$\kappa(K)$$ instead.

# 3 Lecture 1: Discussion and Review

## 3.1 Valuations

• Transcendence bases
• Lüroth problem

For $$K/k$$ a one variable function field, if we want a curve $$C/k$$, what are the points? We’ll use valuations, see NT 2.1. See also completions, residue fields. If $$R \subset K$$ a field, $$R$$ is a valuation ring of $$K$$ if for all $$x\in K^{\times}$$, at least one of $$x, x^{-1} \in R$$.

The valuation rings of $${\mathbb{Q}}$$ are $${\mathbb{Z}}_{(p)}\mathrel{\vcenter{:}}={\mathbb{Z}}[\left\{{{1\over \ell} {~\mathrel{\Big|}~}\ell\neq p}\right\}]$$ for all primes $$p$$.

See also Krull valuations, which take values in some totally ordered commutative group.

Show that a valuation ring is a local ring, i.e. it has a unique maximal ideal.

Where does the log come from?

There is a $$p{\hbox{-}}$$adic valuation: \begin{align*} v: {\mathbb{Q}}&\to {\mathbb{Z}}_{(p)} \\ {a\over b} = p^n {u \over v} &\mapsto n .\end{align*}

Then we recover \begin{align*} {\mathbb{Z}}_{(p)} = \left\{{x\in {\mathbb{Q}}^{\times}{~\mathrel{\Big|}~}v_p(x) \geq 0}\right\} \cup\left\{{0}\right\} \\ {\mathfrak{m}}_{(p)} = \left\{{x\in {\mathbb{Q}}^{\times}{~\mathrel{\Big|}~}v_p(x) > 0}\right\} \cup\left\{{0}\right\} \\ .\end{align*}

There is a $$p{\hbox{-}}$$adic norm \begin{align*} {\left\lvert {{\,\cdot\,}} \right\rvert}_p: {\mathbb{Q}}&\to {\mathbb{R}}\\ 0 & \mapsto 0 \\ x &\mapsto p^{-n} = p^{-v_p(x)} .\end{align*}

Then we get an ultrametric function, a non-archimedean function \begin{align*} d_p: {\mathbb{Q}}^2 \to {\mathbb{R}}\\ (x, y) &\mapsto {\left\lvert {x- y} \right\rvert}_p .\end{align*}

We then recover $$v_p(x) = -\log_p {\left\lvert {x} \right\rvert}_p$$.6

## 3.2 Places

For $$A\subset K$$ a subring of a field, we’ll be interested in the place $$\tilde \Sigma = \left\{{\text{Valuation rings } R_v \text{ of } K}\right\} {~\mathrel{\Big|}~}A \subset R_v \subsetneq K$$. Thus the valuation takes non-negative values on all elements of $$K$$. Can equip this with a topology (the “Zariski” topology, not the usual one). This is always quasicompact, and called the Zariski-Riemann space. Can determine a sheaf of rings to make this a locally ringed space.

We can define an equivalence of valuations and define the set of places \begin{align*} \Sigma(K/k) \mathrel{\vcenter{:}}=\left\{{\text{Nontrivial valuations } v\in K {~\mathrel{\Big|}~}v(x) \geq 0\, \forall x\in k^{\times}}\right\} ,\end{align*} which will be the points on the curve. Here the Zariski topology will be the cofinite topology (which is not Hausdorff). Scheme-theoretically, this is exactly the set of closed points on the curve.

A point $$p\in X$$ a topological space is a generic point iff its closure in $$X$$ is all of $$X$$.

Note we will have unique models for curves, but this won’t be the case for surfaces: blowing up a point will yield a birational but inequivalent surface.

## 3.3 Divisors

The divisor group of $$K$$ is the free $${\mathbb{Z}}{\hbox{-}}$$module on $$\Sigma(K/k)$$

This comes with a degree map \begin{align*} \deg: \operatorname{Div}(K) \to {\mathbb{Z}} \end{align*} which need not be surjective.

Consider the map \begin{align*} \phi_d: K^{\times}&\to \operatorname{Div}(K) \\ f &\mapsto (f) .\end{align*} Then we define $$\operatorname{im}\phi_d$$ as the subgroup of principal divisors.

Define the class group of $$K$$ as \begin{align*} \operatorname{cl}(K) \mathrel{\vcenter{:}}=\left\{{\text{Divisors}}\right\} / \left\{{\text{Principal divisors}}\right\} \mathrel{\vcenter{:}}=\operatorname{Div}(K) / \operatorname{im}\phi_d .\end{align*}

We can define the class group as divisors modulo principle divisors $$\operatorname{cl}(K) = \operatorname{Div}(K) / \operatorname{im}(K^{\times})$$ and the Riemann-Roch space $$\mathcal{L}(D)$$. The Riemann-Roch theorem will then be a statement about $$\dim \mathcal{L}(D)$$.

# 4 Lecture 2: Field Theory Preliminaries

## 4.1 Base Extension

Given some object $$A/k$$ and $$k\hookrightarrow\ell$$ is a field extension, we would like some extended object $$A/\ell$$.

An affine variety $$V/k$$ is given by finitely many polynomials in $$p_i \in k[t_1, \cdots, t_n]$$, and base extension comes from the map $$k[t_1, \cdots, t_n] \hookrightarrow\ell[t_1, \cdots, t_n]$$. More algebraically, we have the affine coordinate ring over $$k$$ given by $$k[V] = k[t_1,\cdots, t_n]/\left\langle{p_i}\right\rangle$$, the ring of polynomial functions on the zero locus corresponding to this variety. We can similarly replace $$k$$ be $$\ell$$ in this definition. Here we can observe that $$\ell[V] \cong k[V] \otimes_k \ell$$.

In general we have a map \begin{align*} {\,\cdot\,}\otimes_k \ell & \\ \left\{{k{\hbox{-}}\text{vector spaces}}\right\} &\to \left\{{\ell{\hbox{-}}\text{vector spaces}}\right\} \\ \left\{{k{\hbox{-}}\text{algebras}}\right\} &\to\left\{{\ell{\hbox{-}}\text{algebras}}\right\} .\end{align*} This will be an exact functor on the category $$k{\hbox{-}}\text{Vect}$$, i.e. $$\ell$$ is a flat module. Here everything is free, and free $$\implies$$ flat, so things work out nicely.

What about for function fields? Since $$k$$ is a $$k{\hbox{-}}$$algebra, we can consider $$k\otimes_k \ell$$, however this need not be a field. Note that tensor products of fields come up very often, but don’t seem to be explicitly covered in classes! We will broach this subject here.

If $$\ell/k$$ is algebraic and $$\ell\otimes_k \ell$$ is a domain, the $$\ell = k$$.

In other words, this is rarely a domain. A hint: start with the monogenic case, and also reduce to the case where the extension is not just algebraic but finite.

Tensor products of field extensions are still interesting: if $$\ell/k$$ is finite, it is galois $$\iff$$ $$\ell \otimes_k \ell \cong \ell^{[\ell: k]}$$. So its dimension as an $$\ell{\hbox{-}}$$algebra is equal to the degree of $$\ell/k$$, so it splits as a product of copies of $$\ell$$.

We’d like the tensor product of a field to be a field, or at least a domain where we can take the fraction field and get a field. This hints that we should not be tensoring algebraic extensions, but rather transcendental ones.

For $$\ell/k$$ a field extension,

1. Show $$k(t) \otimes_k \ell$$ is a domain with fraction field $$\ell(t)$$.

2. Show it is a field $$\iff$$ $$\ell/k$$ is algebraic.

Let $$k_1, k_2 / k$$ are field extensions, and suppose $$k_1 \otimes_k k_2$$ is a domain. Then this is a field $$\iff$$ at least one of $$k_1/k$$ or $$k_2/k$$ is algebraic.7

So we’ll concentrate on when $$K \otimes_k \ell$$ is a domain.

## 4.2 When Extensions Preserve Being a Domain

What’s the condition on a function field $$K/k$$ that guarantees this, i.e. when extending scalars from $$k$$ to $$\ell$$ still yields a domain?

If this remains a domain, we’ll take the fraction field and call it the base change.

If $$K/k$$ is finitely generated (i.e. a function field) and $$K\otimes_k \ell$$ is a domain, then $$ff(K\otimes_k \ell)/\ell$$ is finitely generated.

The point here is that if taking a function field and extending scalars still results in a domain, we’ll call the result a function field as well. Most of all, we want to base change to the algebraic closure. We’ll have issues if the constant field is not just $$k$$ itself:

If $$K\otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ is a domain, then the constant field $$\kappa(K) = k$$.

Use the fact that $${\,\cdot\,}\otimes_k V$$ is exact. We then get an injection Here we use the injections $$\kappa(K) \hookrightarrow\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ and $$\kappa(K) \hookrightarrow K$$.

We now have an injection of $$k{\hbox{-}}$$algebras, and subrings of domains are domains. So apply the first exercise: the only way this can happen is if $$\kappa(K) = k$$.

Describe $${\mathbb{C}}(t) \otimes_{\mathbb{R}}{\mathbb{C}}$$, tensored as $${\mathbb{R}}{\hbox{-}}$$algebras.

Won’t be a domain by the lemma, and will instead be some $${\mathbb{C}}(t){\hbox{-}}$$algebra of dimension 2.

## 4.3 Good Base Change For Function Fields

In order to have a good base change for our function fields, we want to constant extension to be trivial, i.e. $$\kappa(K) = k$$. This requires that the ground field be algebraically closed.

In this case, you might expect that extending scalars to the algebraic closure would yield a field again. This is true in characteristic zero, but false in positive characteristic.

If $$\kappa(K) = k$$, must $$K\otimes_K \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ be a field?

If that’s true and we’re in positive characteristic, recalling the for an algebraic extension this being a field is equivalent to it being a domain. But if that’s a domain, the tensor product of every algebraic extension must be a domain, which is why this is an important case.

If so, then $$K\otimes_k k^{1\over p}$$ is a field, where $$k^{1\over p} \mathrel{\vcenter{:}}= k\qty{\left\{{x^{1\over p} {~\mathrel{\Big|}~}x\in k }\right\}}$$ is obtained by adjoining all $$p$$th roots of all elements. This is a purely inseparable extension. The latter condition (this tensor product being a field) is one of several equivalent conditions for a field to be separable.8

Recall that $$K/k$$ is transcendental, and there is an extended notion of separability for non-algebraic extensions. Another equivalent condition is that every finitely generated subextension is separably generated, i.e. it admits a transcendence basis $$\left\{{x_i}\right\}$$ such that $$k\hookrightarrow k(\left\{{x_i}\right\}) \hookrightarrow F$$ where $$F/k(\left\{{x_i}\right\})$$ is algebraic and separable. Such a transcendence basis is called a separating transcendence basis. Since we’re only looking at finitely generated extensions, we wont’ have to worry much about the difference between separable and separably generated.

What’s the point? There’s an extra technical condition to ensure the base change is a field: the function field being separable over the ground field. Is this necessarily the case if $$\kappa(K) = k$$?

No, for fairly technical reasons.

Set $$k = {\mathbb{F}}_p(a, b)$$ a rational function field in two variables as the ground field. Set \begin{align*} A \mathrel{\vcenter{:}}= k[x, y]/ \left\langle{ax^p + b-y^b}\right\rangle .\end{align*} Then $$A$$ is a domain, so set $$k = ff(A)$$.

$$\kappa(K) = k$$, so $$k$$ is algebraically closed in this extension, but $$K/k$$ is not separable.

How to show: extending scalars to $$k^{1\over p}$$ does not yield a domain.

Let $$\alpha, \beta \in \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ such that $$\alpha^p = a, \beta^b = b$$, so \begin{align*} ax^p + b-y^b = (\alpha x + \beta - y)^p ,\end{align*} which implies $$K \otimes_k k^{1\over p}$$ is not a domain: $$k[x, y]$$ is a UFD, so the quotient of a polynomial is a domain iff the polynomial is irreducible. However, the $$p$$th power map is a homomorphism, and this exhibits the image of the defining polynomial as something non-irreducible.

Note that $$f(x, y) = ax^p + b - y^p$$ is the curve in this situation. The one variable function field is defined by quotienting out a function in two variables and taking the function field. Every 1-variable function field can be obtained in this way. Therefore this polynomial is irreducible, but becomes reducible over the algebraic closure. So we’d like the polynomial to be irreducible over both.

This is pretty technical, but we won’t have to worry if $$k = k^{1\over p}$$. Equivalently, frobenius is surjective on $$k$$, i.e. $$k$$ is a perfect field.

If $$k$$ is not perfect, it can happen (famous paper of Tate) making an inseparable base extension can decrease the genus of the curve.

Recall that the perfect fields are given by:

• Anything characteristic zero, every reducible polynomial is separable.
• Any algebraically closed field
• Finite fields (frobenius is always injective)

Imperfect fields include:

• Function fields in characteristic $$p$$
• Complete discretely valued fields $$k((t))$$ in characteristic $$p$$9

For field extensions $$K/k$$, TFAE

1. $$\kappa(K) = k$$ and $$K/k$$ is separable

2. $$K\otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ is a domain, or equivalently a field

3. For all field extensions $$\ell/k$$, $$K\otimes_k \ell$$ is a domain.

Note that this allows making not just an algebraic base change, but a totally arbitrary one.

A field extension satisfying these conditions is called regular.

Regular corresponds to “nonsingular” in this neck of the woods. The implication $$2\implies 3$$ is the interesting one. To prove it, reduces to showing that if $$k= \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ and $$R_i$$ are domains that are finitely generated as $$k{\hbox{-}}$$algebras, then $$R_1 \otimes_k R_2$$ is also a domain. This doesn’t always happen, e.g. $${\mathbb{Q}}(\sqrt{2}) \otimes_{\mathbb{Q}}{\mathbb{Q}}(\sqrt{2})$$ is not a domain. Really need algebraically closed.

This is a result in affine algebraic geometry. An algebra that is a domain and finitely generated over a field is an affine algebraic variety, more precisely it is integral. The tensor product on the coordinate ring side corresponds to taking the product of varieties. Thus the fact here is that a product of integral varieties remains integral, as long as you’re over an algebraically closed field. Proof uses Hilbert’s Nullstellensatz.

1. Show that $$k(t) / k$$ is regular.10

2. Show every purely transcendental extension is regular.

3. Show that for a field $$k$$, every extension is regular $$\iff$$ $$k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$.

4. Show $$K/k$$ is regular $$\iff$$ every finitely generated subextension is regular.

## 4.4 Example of a Non-Regular Family of Function Fields

Choose an elliptic curve $$E/{\mathbb{Q}}(t)$$ with $$j{\hbox{-}}$$invariant $$t$$. For $$N\in {\mathbb{Z}}^{+}$$, define $$\tilde K_N \mathrel{\vcenter{:}}={\mathbb{Q}}(t)(E[N])$$ the $$N{\hbox{-}}$$torsion field of $$E$$. Then $$\tilde K_N/{\mathbb{Q}}(t)$$ is a finite galois extension with galois group isomorphic to the image of the modular galois representation11

\begin{align*} \rho_N: g({\mathbb{Q}}(t)) \to \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) \pmod N .\end{align*}

$$\rho_N$$ is surjective, and \begin{align*} {\operatorname{Aut}}(\tilde K_N / {\mathbb{Q}}(t)) \cong \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) .\end{align*}

$$\det \rho_N = \chi_N \pmod N$$, the cyclotomic character, and therefore $$\chi_N$$ restricted to $$g(\tilde K_N)$$ is trivial, so $$\tilde K_N \supset {\mathbb{Q}}(\zeta_N)$$. For $$N\geq 3$$, $${\mathbb{Q}}(\zeta_N) \supsetneq {\mathbb{Q}}$$, so $$\tilde K_N / {\mathbb{Q}}(t)$$ is a non-regular function field.

Actually $$\tilde K_N$$ depends on the choice of $$E$$: difference choices of nonisomorphic curves with the same $$j{\hbox{-}}$$invariant differ by a quadratic twist and the $$\rho_N$$ differ by a quadratic character on $$g({\mathbb{Q}}(t))$$. Importantly, this changes the kernel, and thus the field.

To fix this, we look at the reduced galois representation, the following composition: \begin{align*} \mkern 1.5mu\overline{\mkern-1.5mu\rho\mkern-1.5mu}\mkern 1.5mu_N: g({\mathbb{Q}}(t)) \to \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) \twoheadrightarrow\operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}})/\left\{{\pm I}\right\} .\end{align*}

We obtain a field theory diagram

So if you just take the field fixed by $$\pm I$$, you get $$K_N$$. In this case, the reduced galois representation depends only on the $$j{\hbox{-}}$$invariant, and not on the model chosen. So the function field $$K_N/{\mathbb{Q}}(t)$$ is the “canonical” choice.

Does this make $$K_N/{\mathbb{Q}}(t)$$ regular?

No, $$\rho_N(g(K_N)) = \left\{{\pm I}\right\}$$ and $$\det(\pm I) = 1$$, so we still have $$K_N \supset {\mathbb{Q}}(\zeta_N)$$.

In this course, we’ll identify algebraic curves over $$k$$ and one-variable function fields $$K/k$$. The function field $$K_N$$ corresponds to an algebraic curve $$X(N)/{\mathbb{Q}}$$ that is “nicer” over $${\mathbb{Q}}(\zeta_N)$$. In fact, see Rohrlich: $$\kappa(K_N) = {\mathbb{Q}}(\zeta_N)$$. Our curves will have points (equal to valuations) which will have degrees. If the constant subfield is not just $$k$$, this prevents degree 1 points on the curve. By Galois theory, for every subgroup $$H \subseteq \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) / \left\{{\pm I}\right\}$$, we’ll get a function field $${\mathbb{Q}}(H) \mathrel{\vcenter{:}}= H_N^H$$.gg In this case, $${\mathbb{Q}}(H)/{\mathbb{Q}}$$ is regular $$\iff$$ $$\det(H) = ({\mathbb{Z}}/N{\mathbb{Z}})^{\times}$$.

Later we’ll understand the residues at points as the residue fields of some DVRs, then the residue field will always contain the field of constants.

# 5 Lecture 3: Last of Preliminaries

Today we’ll be wrapping up the last of the preliminaries. Upcoming: one-variable function fields and their valuation rings.

## 5.1 Polynomials Defining Regular Function Fields

Where’s the curve in all of this?

This will come from an equation like $$f(x, y) = 0$$.

Let $$R_1, R_2$$ be $$k{\hbox{-}}$$algebras that are also domains with fraction fields $$K_i$$. Show $$R_1 \otimes_k R_2$$ is a domain $$\iff$$ $$K_1 \otimes_k K_2$$ is a domain.12

## 5.2 Geometric Irreducibility

A polynomial of positive degree $$f\in k[t_1, \cdots, t_n]$$ is geometrically irreducible if $$f\in \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[t_1, \cdots, t_n]$$ is irreducible as a polynomial.

If $$n=1$$ then $$f$$ is geometrically irreducible $$\iff$$ $$f$$ is linear, i.e. of degree 1. Let $$f$$ be irreducible, then since polynomial rings are UFDs then $$\left\langle{f}\right\rangle$$ is a prime ideal (irreducibles generate principal ideals) and $$k[t_1, \cdots, t_n]/\left\langle{f}\right\rangle$$ is a domain. Let $$K_f$$ be the fraction field.

1. Above for $$1\leq i \leq n$$ let $$x_i$$ be the image of $$t_i$$ in $$K_f$$. Show that $$K_f = k(x_1, \cdots, x_n)$$.

2. Show that if $$K/k$$ is generated by $$x_1, \cdots, x_n$$, then it is the fraction field of $$k[t_1, \cdots, t_n] /{\mathfrak{p}}$$ for some prime ideal $${\mathfrak{p}}$$ (equivalently, a height 1 ideal).

Suppose that $$f$$ is geometrically irreducible.

1. The function field $$K/k$$ is regular.

2. For all $$\ell/k$$, $$f\in \ell[t_1, \cdots, t_n]$$ is irreducible.

In this case we say $$f$$ is absolutely irreducible as a synonym for geometrically irreducible.

By definition of geometric irreducibility, $$\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[t_1, \cdots, t_n]/\left\langle{f}\right\rangle = k[t_1, \cdots, t_n]/\left\langle{f}\right\rangle \otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ is a domain. The exercise shows that $$K_f \otimes_k k$$ is a domain, so $$K_f$$ is regular. It follows that for all $$\ell/k$$, $$K_f \otimes_k \ell$$ is a domain, so $$\ell[t_1, \cdots, t_n]/\left\langle{f}\right\rangle$$ is a domain.

Geometrically irreducible polynomials are good sources of regular function fields.

Let $$k$$ be a field, $$d\in {\mathbb{Z}}^+$$ such that $$4\nmid d$$ and $$p(x) \in k[x]$$ be positive degree. Factor $$p(x) = \prod_{i=1}^r (x-a_i)^{\ell_i}$$ in $$\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[x]$$.

1. Suppose that for some $$i$$, $$d\nmid\ell_i$$. Show that $$f(x, y) \mathrel{\vcenter{:}}= y^d - p(x) \in k[x, y]$$ is geometrically irreducible. Conclude that $$K_f \mathrel{\vcenter{:}}= ff\qty{k[x, y] / \left\langle{y^d - p(x)}\right\rangle}$$ is a regular one-variable function field over $$k$$, and thus elliptic curves yield regular function fields.13

2. What happens when $$4\divides d$$?

Assume $$k$$ is a field, if necessary assuming $$\operatorname{ch}(k) \neq 2$$.

1. Let $$f(x, y) = x^2 - y^2 -1$$ and show $$K_f$$ is is rational: $$K_f = k(z)$$.

2. Let $$f(x, y) = x^2 + y^2 - 1$$. Show that $$K_f$$ is again rational.

3. Let $$k = {\mathbb{C}}$$ and $$f(x, y) = x^2 + y^2 + 1$$, $$K_f$$ is rational.

4. Let $$k= {\mathbb{R}}$$. For $$f(x ,y) = x^2 + y^2 + 1$$, is $$K_f$$ rational?14

Can we always construct regular function fields using geometrically irreducible polynomials?

In several variables, no, since not every variety is birational to a hypersurface. In one variable, yes, as the following theorem shows:

## 5.3 Our Function Fields are Geometrically Irreducible

Let $$K/k$$ be a one variable function field (finitely generated, transcendence degree one). Then

1. If $$K/k$$ is separable, then $$K = k(x, y)$$ for some $$x, y\in K$$.

2. If $$K/k$$ is regular (separable + constant subfield is $$k$$, so stronger) then $$K \cong K_f$$ for a geometrically irreducible $$f\in k[x ,y]$$.

Recall separable implies there exists a separating transcendence basis.

This means there exists a primitive element $$x\in K$$ such that $$K/k(x)$$ is finite and separable. By the Primitive Element Corollary (FT 7.2), there exist a $$y\in K$$ such that $$K = k(x, y)$$.

Omitted for now, slightly technical.

Importance of last result: a regular function field on one variable corresponds to a nice geometrically irreducible polynomial $$f$$.

Note that the plane curve module may not be smooth, and in fact usually is not possible. I.e. $$k[x ,y]/\left\langle{f}\right\rangle$$ is a one-dimensional noetherian domain, which need not be integrally closed.

Can every one variable function field be 2-generated?

Yes, as long as the ground field is perfect. In positive characteristic, the suspicion is no: there exists finite inseparable extensions $$\ell/k$$ that need arbitrarily many generators. However, what if $$K/k$$ has constant field $$k$$ but is not separable? Riemann-Roch may have something to say about this.

\begin{align*} ax^p + b - y^b \end{align*}

We can find examples of nice function fields by taking irreducible polynomials in two variables. This will define a one-variable function field. If the polynomial is geometrical reducible, this produces regular function fields.

# 6 Lecture 4: Chapter 1, One Variable Function Fields and Their Valuations

Since we have the field-theoretic preliminaries out of the way, we now start studying one-variable function fields in earnest. The main technique that we use to extract the geometry will be the theory of valuations. These may be familiar from NTII, but we will cover them in more generality here.

## 6.1 Valuation Rings and Krull Valuations

Recall that NTII approach to valuations:

A valuation on a field $$K$$ is a map $$v:K\to {\mathbb{R}}\cup\left\{{\infty}\right\}$$ such that $$v(K^{\times}) \subset {\mathbb{R}}$$, $$v(0) = \infty$$, and $$v$$ is of the form $$-\log({\left\lvert {{\,\cdot\,}} \right\rvert})$$ where $${\left\lvert {{\,\cdot\,}} \right\rvert}: K \to [0, \infty)$$ is an ultrametric norm.15 Recall that an ultrametric norm satisfies not only the triangle inequality but the ultrametric triangle inequality, i.e. $$d(x, z) \leq \max(x, z)$$.

We now take an algebraic approach to this definition, where we’ll end up replacing $${\mathbb{R}}$$ with something more general.

A subring $$R$$ of a field $$K$$ is a valuation ring if for all $$x\in K^{\times}$$, at least one of $$x$$ or $$x^{-1}$$ is in $$R$$.

This is a “largeness” property. It also implies that $$K = \operatorname{ff}(R)$$.

## 6.2 Group of Divisibility

Given any integral domain $$R$$ with fraction field $$K$$, the group of divisibility $$G(R)$$ is defined as the partially ordered commutative group16

\begin{align*} G(R) \mathrel{\vcenter{:}}= K^{\times}/ R^{\times} .\end{align*} We will write the group law here additively. The ordering is given by $$x\leq y \iff y/x \in R$$.

Note that the way the partial order is written, it’s a relation on $$K^{\times}$$, but it is not quite a partial ordering there. It is reflexive and transitive, but need not be antireflexive: if $$x/y, y/x\in R$$ then $$x,y$$ differ by an element of $$u\in R^{\times}$$ so that $$x=uy$$. In particular, they need not be equal. This gives a structure of a quasiordering, and if you set $$x\sim y \iff x\leq y$$ and $$y\leq x$$, this leads to an equivalence relation, and modding out by it yields a partial order. Here this is accomplished by essentially trivializing units.

Another way to think of $$G(R)$$ is as the nonzero principal fractional ideals of $$K$$, since any two generators of an ideal will differ by a unit.

Inside this group there is a positive cone $$G(R)^+$$ of elements that are “nonnegative”: since we’re in a commutative setting, the zero element is equal to 1, and the positive cone is given by $$\left\{{y\geq 0}\right\} = \left\{{y\in R}\right\}$$, and is thus given by the group $$G(R)^+ = (R, \cdot)$$.

This is very general: if you’re studying factorization in integral domains, many properties are reflected in $$G(R)$$. E.g. being a UFD (the most important factorization property!) implies that $$G(R)$$ is a free commutative group.

In general this is only a partially ordered group and not totally ordered. For example, take $$R = {\mathbb{Z}}$$ and $$x=2, y=3$$, then neither of $$2/3, 3/2$$ are in $${\mathbb{Z}}$$, so $$x\not\leq y$$ and $$y\not\leq x$$. On the other hand, if we do have a total order, then either $$x$$ or $$x^{-1}$$ is in the ring, which are exactly valuation subring of a field.

$$R$$ is a valuation ring $$\iff$$ $$G(R)$$ is totally ordered.

Note that $${\mathbb{R}}$$ is a totally ordered group.

## 6.3 Generalized Valuations

This makes $$G(R)$$ the “target group” of a generalized analytic valuation. Whenever we have a valuation ring, we have a totally ordered commutative group, and the valuation $$v: K^{\times}\to G(R)$$ is a quotient map which we can extend to $$K$$ by $$v(0) \mathrel{\vcenter{:}}=\infty$$. This has some familiar properties:

• (VRK1) For all $$x,y\in K^{\times}$$,17

\begin{align*} v(xy) = v(x) + v(y) .\end{align*}

• (VRK2) For all $$x,y \in K^{\times}$$ such that $$x+y\neq 0$$, \begin{align*} v(x+y) \geq \min(v(x), v(y)) .\end{align*}

For ultrametric norms, all triangles are isosceles: is that true for this type of function? The answer is yes, by the following exercise:

If $$v(x) \neq v(y)$$, then $$v(x+y) = \min(v(x), v(y))$$.

So the properties here are formally identical to the NTII notion of valuation, with $$({\mathbb{R}}, +, \leq)$$ replaced by $$(G(R), +, \leq)$$.

Conversely, if $$v: K^{\times}\to G$$ is a map into a totally ordered commutative group satisfying VRK1 and VRK218, then \begin{align*} R_v \mathrel{\vcenter{:}}=\left\{{x\in K^{\times}{~\mathrel{\Big|}~}v(x) \geq 0}\right\} \cup\left\{{0}\right\} \end{align*} is a valuation ring.19 We can thus extract valuation rings in this situation.

A valuation ring is local, i.e. there is a unique maximal ideal \begin{align*} \mathfrak{m}_v \mathrel{\vcenter{:}}=\left\{{x\in K^{\times}{~\mathrel{\Big|}~}v(x) > 0}\right\} \cup\left\{{0}\right\} .\end{align*}

These two constructions are morally mutually inverse. This doesn’t hold on the nose, since there is extraneous data in the new analytic valuation. Recall that in NTII we have a notion of equivalence of norms, and two distinct norms that are equivalent can give rise to the same valuation. For example, given a valuation, one can scale it by $$\alpha \in {\mathbb{R}}$$, and it’s easy to check that this gives the same valuation. It is possible for the valuation not to surject onto $${\mathbb{R}}$$, but this doesn’t happen in practice. The image is usually infinite cyclic, what we call a discrete valuation, and so one is led to the definition of the value group of the valuation as its image. If you have a notion of equivalence of Krull valuations, you want to allow for isomorphisms of the value group. The cleanest notion of equivalence is thus the following:

Two Krull valuations on a field $$K$$ are equivalent iff their valuation rings are equal.

Going back to NTII, if you have two nonarchimedean norms on a field, then there are many equivalent conditions for equivalence, and this is one of them.

Some general valuation theory:

• Every totally ordered commutative group is a group of divisibility.20

• A totally ordered group has rank 1 if it is nontrivial and embeds into $${\mathbb{R}}$$

• If the value group is trivial, $$R = K$$
• A Krull valuation of rank at most 1 is the NTII notion of a valuation.

For $$n\geq 2$$, put the lexicographic order on $${\mathbb{Z}}^n$$, and show this has rank strictly larger than 1. Thus $${\mathbb{Z}}^n\hookrightarrow{\mathbb{R}}$$ as a commutative group, but not as a totally ordered commutative group.

In fact, for any ordered group $$G$$, one can attach a rank: a cardinal number $$r(G)$$. Here, $$r(({\mathbb{Z}}^n, \text{lex})) = n$$. This is useful when studying $$\operatorname{Spec}(R)$$ for $$R$$ a DVR.

A valuation of rank bigger than 1 does not induce a norm on $$K$$ in the metric sense, although this is not so important. A closer notion is expanding the notion of a metric space by allowing the metric to be defined on $$X$$ as $$d: X\times X \to R$$ for some $$R$$ more general than $${\mathbb{R}}$$, like a totally ordered group or a nonarchimedean field. This would yield a class of topological spaces that are reminiscent of metric spaces.

## 6.4 Regular or Centered Valuations

Let $$v:K^{\times}\to (G, +)$$ be a Krull valuation and let $$A \subset K$$ be a subring of $$K$$. Then $$v$$ is $$A{\hbox{-}}$$regular or centered in $$A$$ if $$A$$ is a subset of some valuation ring $$R_v$$. In this case, $$\mathfrak{p} \mathrel{\vcenter{:}}=\mathfrak{m}_v \cap A \in \operatorname{Spec}(A)$$ is denoted the center of $$v$$ in $$A$$.21

The term regularity here arises because we’ll want to think of elements of $$A$$ as functions and the valuation as a type of point, then the notion of being a regular function at a point will carry over. The center is the subset of $$A$$ with strictly positive valuation. Also recall that pulling back prime ideals yields prime ideals, and maximal ideals are a special kind of prime ideal, but in general pulling back a maximal ideal may not result in another maximal ideal. So somehow the valuation affects every subring on which it is regular.

For $$A \subset K$$, define \begin{align*} \Sigma(K/A) &\mathrel{\vcenter{:}}=\left\{{\text{valuation rings } A \subset R \subsetneq K {~\mathrel{\Big|}~}K = \operatorname{ff}(R)}\right\} \\ \tilde \Sigma(K/A) &\mathrel{\vcenter{:}}=\left\{{\text{valuation rings } A \subset R \subseteq K {~\mathrel{\Big|}~}K = \operatorname{ff}(R)}\right\} .\end{align*} The set $$\tilde \Sigma(K/A)$$ is the Zariski-Riemann space.

Note that in this definition, we’re taking all $$A{\hbox{-}}$$regular valuation rings $$R$$ in $$K$$. If someone says $$R$$ is a valuation ring of $$K$$, they likely mean that $$K = \operatorname{ff}(R)$$. Note that fields are valuation rings, so otherwise, any subfield of $$K$$ would also be a valuation ring of $$K$$. Here, $$K$$ itself plays the role of a generic point. (?) The only difference in these two definitions is that in the first, the trivial valuation ring is being excluded.

If $$K/k$$ is a one variable function field22 , then $$\Sigma(K/k)$$ will be the points of the associated algebraic curve or places. These can be thought of as valuation rings, or equivalence classes of Krull valuations, where two valuations are equivalent if they have the same valuation ring.

In terms of scheme theory, these will be the closed points of our algebraic curve. We will view elements $$f\in K$$ as meromorphic functions on $$\Sigma(K/k)$$.

## 6.5 Topological Considerations

The Zariski topology on $$\Sigma(K/A)$$ has a sub-base \begin{align*} \left\{{U(f) {~\mathrel{\Big|}~}f\in K }\right\} && U(f) \mathrel{\vcenter{:}}=\left\{{v\in \tilde \Sigma(K/A) {~\mathrel{\Big|}~}v(f) \geq 0}\right\} = \tilde \Sigma(K/ A[f]) .\end{align*} and we thus take the minimal topology such that all of these sets are open. In other words, every open set is a finite intersection and/or arbitrary unions, including empty intersections/unions. The last term is precisely the subring generated by $$A$$ and $$f$$. Thus a base is $$U(f_1, \cdots, f_n) = \tilde \Sigma(K / A[f_1, \cdots, f_n])$$. The Zariski topology on $$\Sigma(K/A)$$ is defined the same way and/or via the subspace topology, since this removes a single point.

We thus get the subrings of $$K$$ that contain $$A$$ and are finitely generated as $$A{\hbox{-}}$$algebras. We’ll be specifically looking at the case where $$A$$ is a field and $$K$$ is a one variable function field.

$$\tilde \Sigma(K/A)$$ is quasi-compact.

See Mazamara (?) in the chapter discussing valuation rings.

Note that by definition, $$v_n \not\in \Sigma(K/A)$$. In $$\tilde \Sigma(K/A)$$, we have a trivial valuation $$v_n$$ whose value group is trivial and valuation ring is $$K$$ itself, and $$v_n$$ is a generic point of $$\Sigma(K/A)$$: its closure is the entire space. In other words, it is in every nonempty open subset. Since we have at least one generic point, and in general there may be many, if $${\left\lvert {\tilde\Sigma(K/A) > 1} \right\rvert}$$ then this is not a separated ($$T_1$$) space since the point is not closed.23 Another example of such a space would be $$\operatorname{Spec}(R)$$ for $$R$$ a commutative ring with positive Krull dimension, which will be Kolmogorov ($$T_0$$) but not separated. Such a spectrum is the underlying topological space of some affine scheme, and in general, schemes will have these kinds of properties that are bad (but not too bad).

In our case of interest, when $$K/k$$ is finitely generated of transcendence degree one, we’ll see that this is the cofinite topology on an infinite space: the proper closed subsets are precisely the finite subsets, or equivalently every nonempty open subset has finite complement. This is far from Hausdorff: the intersection of two open subsets will still have finite complement, so any two nonempty open subsets must intersect.

It’s not generally true that just removing the generic point $$v_n$$ will make the space separated, but in our case, it will be. So if we restrict to nontrivial valuation rings, then the underlying set will be infinite and we’ll get the cofinite topology. This will be the coarsest separated topology, i.e. if you want singletons to be closed, finite subsets must be closed. If $$k \subset A \subset K$$ where $$A$$ is a Dedekind domain with fraction field $$K$$, we will see that if we consider not the $$k{\hbox{-}}$$regular elements but the $$A{\hbox{-}}$$regular ones, we’ll get $$\Sigma(K/A) = {\operatorname{maxSpec}}(A)$$ and both Zariski topologies are cofinite. Note that in a Dedekind domain, trading in a prime spectrum for a max spectrum is removing a generic point, so this matches up. The moral: the topology of $$\Sigma(K/k)$$ is not doing anything interesting and we won’t need it much.

## 6.6 Scheme Theory, Resolution of Singularities

When $$K/k$$ instead has transcendence degree bigger than 1, then $$\tilde \Sigma(K/k)$$ is much more interesting. If we were doing things scheme-theoretically, we could try to define a structure sheaf: attaching a sheaf whose stalks are local commutative rings to make it a locally ringed space.24 Here, the choice of a ring is straightforward: literally $$\tilde \Sigma(A, A[f_1, \cdots, f_n])$$. There’s an exercise that shows that although defining a sheaf on the entire space is somewhat annoying, defining it on a basis suffices.

Endow $$\tilde \Sigma(K/k)$$ with the structure of a locally ringed space.

In dimension 1 (the case we’re studying), the corresponding Zariski-Riemann space will be the scheme associated to the complete nonsingular model of the curve. So this valuation-theoretic approach will take you from the function field back to a nice model of the scheme itself. But note that in larger dimensions, there is no unique complete nonsingular model – for example, you can blow any one up to get another – so this pattern can’t possibly continue to hold. In fact, it’s not clear if we even know if there’s one such model!

Thus in dimension $$>1$$, you get something that is decidedly not a scheme, but is still relevant to the study of resolution of singularities for your function field. Where do these come up? Zariski used $$\Sigma(K/A)$$ to prove resolution of singularities25 in characteristic zero and dimensions 2 and 3 in 1944, although dimension 2 was classical by the Italian school. Later, Hironaka (1984) got the Fields medal for proving resolution of singularities for all dimensions in characteristic zero using an ingenious inductive argument that avoided Zariski-Riemann spaces entirely. It remarkably doesn’t use any new objects/tools, just uses existing ones in a clever way. So why talk about Zariski-Riemann spaces at all? In the last 10 years or so, work of Ternkin and Conrad has revived and generalized them. They study relative such spaces.

In positive characteristic, resolution of singularities is only known up to dimension $$\leq 3$$.

## 6.7 Intermediate Rings

The following is an extremely important result from commutative algebra:

Let $$A \subset K$$ be a subring of a field, then \begin{align*} \cap_{v\in \tilde \Sigma(K/A)} R_v ,\end{align*} the intersection of all valuation subrings of the field, is the integral closure of $$A$$ in $$K$$.

The proof is mostly a Zorn’s lemma type of argument. Note that each $$R_v$$ is generally big, contains $$A$$, and $$\operatorname{ff}(R_v) = K$$. Moreover, each valuation ring is integrally closed, although we haven’t proved this yet.

For $$K/k$$ function field, $$\cap_{v\in \Sigma(K/k)} R_v = \kappa(K)$$, the constant subfield of $$K$$.

Note that $$\kappa(K)$$ is the integral (algebraic) closure of $$k$$ in $$K$$. Applying the theorem directly almost works, except the theorem involves $$\tilde \Sigma$$. Can we remove the tilde? Suppose not, this can only happen if $$\Sigma(K/k) = \emptyset$$ and the intersection is just $$K$$ itself, the largest thing in the intersection. But can the integral closure of $$k$$ in $$K$$ be $$K$$ itself? No, since the transcendence degree of the function field is positive. So $$K/k$$ is transcendental, while $$\kappa(K) / k$$ is an algebraic extension, a contradiction.

Note that $$\Sigma(K/k)$$ is nonempty: there is a nontrivial valuation ring between $$k$$ and $$K$$ in great generality, and there are often many.

If $$\operatorname{trdeg}(K/k) = 1$$, then every $$v\in \Sigma(K/k)$$ is discrete and thus has value group isomorphic to $${\mathbb{Z}}$$.

So despite the fact that we’ve introduced a more general notion of higher rank valuations, in dimension 1, every single valuation is discrete.

Let $$v\in \Sigma(K/k)$$ be a place, so its a valuation ring with fraction field $$K$$ that is not $$K$$, then $$R_v$$ is not a field. So its maximal ideal $$\mathfrak{m}_v$$ is nonzero, so choose a nonzero element $$t\in \mathfrak{m}_v$$. Then $$t \in R_v$$ and $$R_v$$ contains $$k$$, so $$k[t] \subset R_v$$. Note that $$k[t]$$ is a PID sitting inside a valuation ring. So restrict this maximal ideal down: $$\mathfrak{m}_v \cap k[t]$$ is a prime ideal of $$k[t]$$ containing $$t$$, and thus the center $$\mathfrak{m}_v \cap k[t] = \left\langle{t}\right\rangle$$. This follows because a prime ideal in the polynomial ring $$k[t]$$ which contains $$t$$ is necessarily generated by $$t$$, since there’s exactly one such ideal.

Now restricting the valuation on $$K$$ to $$k(t) \subset K$$, $$K / k(t)$$ will be a finite extension (from the first lecture). We know $$k(t) \subset K$$, and we can now check that $${ \left.{{v}} \right|_{{k(t)}} }$$ is the $$t{\hbox{-}}$$adic valuation $$v_t$$. Note that $$\mathfrak{m}_v$$ can not contain any other monic irreducible polynomials, since distinct such polynomials are coprime. Since we’re in a PID, this ideal would contain any linear combination of them and thus contain 1. So consider the map \begin{align*} k[t] \hookrightarrow R_b \to G(R_v) = K^{\times}/ R^{\times} .\end{align*} Note that the units of $$k[t]$$ map trivially, using the fact that any element in $$k[t]$$ can be written as $$u \prod p_i^{a_i}$$ with the $$p_i$$ monic irreducible polynomials. The unit maps to zero, along with all of the other monic irreducibles, and thus the image is determined entirely by the image of powers of $$t$$. This whole term goes to zero unless some $$p_i\mapsto t$$, in which case it maps to some power of $$t$$.

So suppose $$t\mapsto \gamma \neq 0\in G(R)$$, which is nonzero because $$t$$ was not a unit (since it was in the maximal ideal). Then the image is exactly $$\gamma^{\mathbb{N}}$$, the non-negative integer powers of the image of $$t$$. But if we know goes on this domain, taking denominators shows where it goes on the fraction field (of a UFD), so the image is the cyclic group generated by $$\gamma$$, i.e. the powers of $$t$$ are literally the only valuations we get. So the image of $$k(t)^{\times}$$ in $$G(R_v)$$ is $$\gamma^{\mathbb{Z}}$$, yielding a discrete valuation. This proves that the restriction to the rational function field $$k(t)$$ is discrete, and we want to use this to deduce that the original valuation is discrete.

We can now use NTII:26 since $$K/k(t)$$ is finite, it follows that $$v$$ is discrete iff $${ \left.{{v}} \right|_{{k(t)}} }$$ is discrete, and thus $$v$$ is discrete.

## 6.8 Valuations of Every Rank

So every place of $$K/k$$ is a discrete valuation as long as the transcendence degree is one, but this is far from the case for degree $$\geq 2$$! In the following example, we’ll have a rational function field, which makes things easier. You need a theory of extending Krull valuations, since we’ll define a non-rank 1 valuation on the rational function field. But an arbitrary finitely generated field extension of degree $$d$$ over $$k$$ is a finite degree extension of the rational function field, and valuation theory will tell you that every valuation downstairs can be extended in full generality to a finite degree field extension, and the rank will not change.

If $$K/k$$ is finitely generated of $$\operatorname{trdeg}\geq 2$$, then $$\Sigma(K/k)$$ has valuations of rank $$d$$.

Note that the Zariski-Riemann space only consists of discrete valuations, which is a characteristic property of one variable function fields. So these higher rank valuations may look weird, but when studying a function field of higher transcendence degree (e.g. for an algebraic surface), these occur.

Let $$k$$ be a field and $$K = k(t_1, \cdots, t_n)$$. Set $$G = {\mathbb{Z}}^n$$ with the lex order, so $$G^{\geq 0} = {\mathbb{N}}^n$$.

• Show that $$k[t_1, \cdots, t_n] = k[G^{\geq 0}]$$, where the RHS is the associated semigroup ring.

• Define $$v: k[G^{\geq 0}]^{\bullet}\to G^{\geq 0}$$ by mapping each polynomial the minimal index of a monomial in its support. For example, \begin{align*} v(a_1 t_1^3 t_2 + a_2 t_1^2 t_2^{10}) = (2, 10) ,\end{align*} which has support $$(3, 1)$$ and $$(2, 10)$$, and we take the min in the lex order.

• Extend $$v$$ to $$v: K^{\bullet}\twoheadrightarrow G$$ satisfying VRK1 and VRK2. Show that $$R_v \mathrel{\vcenter{:}}= v^{-1}(G^{\geq 0}) \cup\left\{{0}\right\}$$ is a valuation ring with value group $$G$$, and in particular, the rank is $$n$$.

Note that doing this for $$n=1$$ reduces to the $$t{\hbox{-}}$$adic valuation, which just keeps track of the smallest power of $$t$$ appearing. Here you can extend to fraction fields by defining $$v(x/y) = v(x) - v(y)$$. The semigroup ring can’t be the valuation ring, since polynomial rings are not local rings, so it’s much bigger. Note also that $${\mathbb{Z}}$$ can be replaced with any group $$G$$, since it’s never used in anything but a psychological fashion.

There is a huge difference between $$\operatorname{trdeg}= 1$$ and $$\operatorname{trdeg}> 1$$, and so we’ll only be working with the former case in this course.

# 7 Lecture 5: Places

An affine domain $$R$$ over a field $$k$$ is a domain that is finitely generated as a $$k{\hbox{-}}$$algebra.

## 7.1 Investigating the Set of Places

We saw an interesting example of a function field in more than one variable which showed that valuations of rank larger than 1 can arise, but this does not happen for one variable function fields. That is, for $$K/k$$ of transcendence degree 1, all valuations on $$K$$ which are trivial on $$k$$ are discrete. We’ll now want to go farther and describe the places $$\Sigma(K/k)$$, which will be the set of points on an algebraic curve. Scheme-theoretically, this will literally be the set of closed points on a certain projective curve whose function field is $$K$$. Note that a priori, finding closed points on a curve over an arbitrary field is hard!

Recall that if $$A$$ is a Dedekind domain such that $$\operatorname{ff}(A) = K$$, then for all $$\mathfrak{p}\in \operatorname{mSpec}(A)$$ there exists a discrete valuation $$v_p$$ on $$K$$. I.e., every maximal ideal induces a discrete valuation that is $$A{\hbox{-}}$$regular, so the valuation ring will contain $$A$$. How is this obtained? Take a nonzero $$x\in K^{\times}$$, and take the corresponding principal fractional ideal $$\left\langle{x}\right\rangle \mathrel{\vcenter{:}}= Ax$$, which we can factor in a Dedekind domain as $$Ax = \prod_{\mathfrak{p} \in \operatorname{mSpec}(A)} \mathfrak{p}^{\alpha_{\mathfrak{p}}}$$ with $$\alpha_{\mathfrak{p}} \in {\mathbb{Z}}$$. This looks like an infinite product, but for any fixed $$x$$, only finitely many $$\alpha$$ are nonzero. Note that these $$\alpha$$ are exactly what we’re looking for: the $$\mathfrak{p}{\hbox{-}}$$adic evaluation of $$x$$ is given precisely by $$v_{\mathfrak{p}}(x) \mathrel{\vcenter{:}}=\alpha_{\mathfrak{p}}$$, where we are using unique factorization of ideals in Dedekind domains. Thus we have a map \begin{align*} v_{{\,\cdot\,}}: \operatorname{mSpec}(A) &\to \Sigma(K/A) \\ \mathfrak{p} &\mapsto v_{\mathfrak{p}} .\end{align*}

So this sends a maximal ideal to a place that is $$A{\hbox{-}}$$regular, and it turns out to be a bijection.

The map $$v$$ is a bijection, and thus we may write \begin{align*} \Sigma(K/A) \cong \operatorname{mSpec}(A) .\end{align*}

$$v$$ is injective.

If $$\mathfrak{p}_1, \mathfrak{p}_2 \in \operatorname{mSpec}(A)$$ are two different maximal ideals. Then there exists an element $$x\in \mathfrak{p}_1 \setminus\mathfrak{p}_2$$, and so $$x^{-1} \in A_{\mathfrak{p}_2} \setminus A_{\mathfrak{p}_1}$$. This follows since if $$x$$ is not in $$\mathfrak{p}_2$$, its $$\mathfrak{p}_2{\hbox{-}}$$adic valuation is zero, and thus the $$\mathfrak{p}_2{\hbox{-}}$$adic valuation of $$x^{-1}$$ is $$-0 = 0$$ as well. On the other hand, since $$x\in \mathfrak{p}_1$$, its $$\mathfrak{p}_1{\hbox{-}}$$adic valuation is positive and therefore $$v_{\mathfrak{p}_1}(x^{-1}) < 0$$ and $$x^{-1}$$ is not in $$A_{\mathfrak{p}_1}$$.

$$v$$ is surjective.

Let $$v\in \Sigma(K/A)$$, so $$A \subset R_v$$, i.e. take a valuation whose valuation ring contains $$A$$. Note that we’re not assuming the valuation is discrete, this can be a general Krull valuation, but we’re trying to show it’s equal to a certain $$p{\hbox{-}}$$adic valuation. As always with a subring of a valuation ring, we can pull back the maximal ideal and consider $$\mathfrak{m}_v \cap A \in \operatorname{Spec}(A)$$. We’re hoping that this is a maximal ideal, since maximals correspond to valuations. Since we’re in a Dedekind domain, the only prime ideal we don’t want this to be is the zero ideal of $$A$$, so suppose it were. Then $$A^{\bullet}\subset R_v^{\times}$$, and so $$K^{\times}\subset R_v^{\times}$$. This is because the only element of the maximal ideal that lies in $$A$$ is zero, so every nonzero element of $$A$$ is not in this maximal ideal and is thus a unit. But for any unit, its inverse is also a unit, yielding the inclusion $$K^{\times}\subset R_v^{\times}$$. The only way this could possibly happen is if $$R_v = K$$, which yields the trivial valuation ring. However, by definition, in $$\Sigma(K/A)$$ we’ve excluded the trivial valuation, so this ideal can not be zero.

So we can conclude that the pullback $$\mathfrak{m}_v \cap A \in \operatorname{mSpec}(A)$$, and so $$A_{\mathfrak{p}} \subset R_v$$. This is from viewing elements in $$A_{\mathfrak{p}}$$ as quotients of elements in $$A$$ whose denominator have $$\mathfrak{p}{\hbox{-}}$$adic valuation zero. Recall that we want to show that $$R_v = A_{\mathfrak{p}}$$. We know $$R_v \subset K$$ is a proper containment, and we can use the fact that a discrete valuation ring is maximal among all proper subrings of its fraction field. In other words, for $$R$$ a DVR, there is no ring $$R'$$ such that $$R \subset R' \subset \operatorname{ff}(R)$$. How do you prove this? This is similar to an early exercise in commutative algebra, where we looked at all rings between $${\mathbb{Z}}$$ and $${\mathbb{Q}}$$, which generalized to looking at all rings between a PID and its fraction field, and a DVR is a local PID. So proving this statement is actually easier.

This is enough to show that $$A_{\mathfrak{p}} = R_v$$, and this $$v\sim v_{\mathfrak{p}}$$.

What the idea? For a general one variable function field $$K/k$$, we’ll produce affine Dedekind domains $$R$$ with $$k \subset R \subset K$$ and $$\operatorname{ff}(R) = K$$. This will give is subrings of this full ring of places that are $$\operatorname{mSpec}$$ of Dedekind domains. How many such domains will we need for their union to be the entire set of places? Just one won’t work, since $$\Sigma(K/k)$$ is like a complete or projective object, and a projective variety of dimension 1 can’t be covered by a single affine variety. However, it turns out that you can always cover it with 2. In fact, if you take any Dedekind domain between $$k$$ and $$\operatorname{ff}(K)$$, the set of missing places (the ones that aren’t regular for any of these domains) will be a nonempty finite set of places. So you can always cover it by finitely many, and two suffices: as a consequence of the Riemann-Roch theorem, after removing any nonempty finite set of places, you’ll have the $$\operatorname{mSpec}$$ of a canonically associated Dedekind domain. We’ll prove this by starting with the case of $$K = k(t)$$.

\begin{align*} {\left\lvert { \Sigma(k(t)/k) \setminus\operatorname{mSpec}k[t] } \right\rvert} = 1 .\end{align*}

Note that $$k \subset k[t] \subset k(t)$$ and $$k[t]$$ is a Dedekind domain, so this fits into the above framework, and moreover we know the maximal ideals of polynomial rings: irreducible monic polynomials. Taking all of these misses exactly one place.

How do we describe this missing place?

## 7.2 Describing the Missing Place

Suppose $$v \in \Sigma(k(t) / k) \setminus\Sigma(k(t) / k[t])$$, so the valuation ring of $$v$$ contains $$k$$ but does not contain $$k[t]$$. Then the valuation ring can not contain $$t$$, and thus $$v(t) < 0$$ and $$v(1/t) = -v(t) > 0$$. Since $$k[1/t]$$ is a PID, so if the valuation wasn’t $$t dash$$regular, it’s $$1/t{\hbox{-}}$$regular by definition. So $$v\in \Sigma(k(t) / k[1/t])$$. Note that $$k[1/t] \cong k[t]$$ as rings. How many valuations on this polynomial ring give positive valuation to $$1/t$$? Exactly one, since this corresponds to a prime ideal, namely $$\left\langle{1/t}\right\rangle$$, so this unique valuation is $$v = v_{1 \over t}$$, the $$1/t{\hbox{-}}$$adic valuation.

That is, if we write $$f\in k(t)$$ as $$(1/t)^n a(1/t)/b(1/t)$$ with $$a, b\in k[t]$$ polynomials with nonzero constant terms, then $$v_{1\over t}(f) = n$$. Note that this process is the same as the one used to compute the $$t{\hbox{-}}$$adic valuation $$v_t$$.

Recall that a valuation on a domain can be uniquely extended to its fraction field by setting $$v(x/y) = v(x) - v(y)$$.

Define $$v_\infty: k(t)^{\times}\to {\mathbb{Z}}$$ by $$p(t)/q(t) \mapsto \deg q - \deg p$$.

1. Show $$v_ \infty \in \Sigma(k(t) / k[1/t])$$.

2. Show $$v_ \infty \sim v_{1\over t}$$ by showing they have the same valuation ring.

3. Show that $$v_ \infty = v_{1\over t}$$.

Note that $$1/t$$ is a uniformizer for $$v_ \infty$$

\begin{align*} \Sigma(k(t) / k) = \operatorname{mSpec}k[t] {\coprod}\left\{{v_ \infty}\right\} .\end{align*}

Note that we know the maximal ideals – the irreducible monic polynomials – but it takes some effort to write them down. If $$k$$ is algebraically closed, however, every such polynomial is linear of the form $$t-\alpha$$ for $$\alpha\in k$$. In this case, $$\operatorname{mSpec}k(t) \cong k$$, and so $$\sigma( \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu (t) / \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu) = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu {\coprod}\left\{{\infty}\right\} = {\mathbb{P}}^1(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu)$$. More generally, the set of places on a rational function field will yield the scheme-theoretic set of closed points on the projective line over $$k$$, which is more complicated if $$k\neq \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ since not all closed points are $$k{\hbox{-}}$$rational. Another way to say this is that if you have a valuation, there is a residue field, and for any place on a one variable function field the residue field will be a finite degree extension of $$k$$. The degree 1 points will be the $$k{\hbox{-}}$$rational points, and so $$\Sigma(k(t) / k)$$ will always contain a copy of $$k$$ but may have closed points of larger degree, making things slightly more complicated. This complication is handled well in both the scheme-theoretic and this valuation-theoretic approach.

## 7.3 Finite Generation in Towers

The next theorem is a fact from commutative algebra:

Let $$A$$ be a domain with $$\operatorname{ff}(A) = K$$. Suppose $$A$$ is a finitely generated $$k{\hbox{-}}$$algebra, let $$L/K$$ be a finite degree field extension, and let $$B$$ be the integral closure of $$A$$ in $$L$$. Then

1. $$B$$ is finitely generated as an $$A{\hbox{-}}$$module.27
1. $$B$$ is an integrally closed domain with $$\operatorname{ff}(B) = L$$ which is finitely generated as a $$k{\hbox{-}}$$algebra.

2. $$\dim A = \dim B$$28

1. If $$A$$ is Dedekind, so is $$B$$.

See Pete’s CA notes sections 18 and 14.

On why these should be true: we have a NTI square We have a domain $$A$$ with a fraction field $$K$$, we take a finite degree extension $$L/K$$, and to complete the square we let $$B$$ be the integral closure of $$A$$ in $$L$$: the collection of elements in $$L$$ satisfying monic polynomials with coefficients in $$A$$.

In our case, we’re additionally assuming that $$A/k$$ is finitely generated as a $$k{\hbox{-}}$$algebra.

On (b): $$B$$ being finitely generated as a $$k{\hbox{-}}$$algebra follows from assuming $$A$$ is, and additionally that $$B$$ is finitely generated as an $$A{\hbox{-}}$$module, and finite generation as a module provides finite generation as an algebra. The result follows from transitivity of finite generation of algebras.

On (c): This is just a property of integral extensions.

On (d): Use the characterization of being Noetherian, integrally closed, and Krull dimension 1. The only thing to check is that $$B$$ is Noetherian, which follows from $$B$$ being finitely generated as a $$k{\hbox{-}}$$algebra and applying the Hilbert basis theorem.

Note that we are not assuming that $$L/K$$ is separable, which is an assumption that would simplify things. By the Krull-Akuzuki theorem, $$B$$ will always be a Dedekind domain, but it need not be finitely generated over $$A$$. So the “stem” to $$k$$ is grounding the situation: it’s not just a Dedekind domain, but rather an affine domain: a domain that is finitely generated over a field. Note that this is much better than an arbitrary Dedekind domain!

## 7.4 Regularity Lemma

Suppose that instead of $$K = \operatorname{ff}(A)$$, we instead have $$A \subset K$$ an arbitrary subring, and $$L/K$$ a finite extension. Taking the integral closure $$B$$ yields another NTI square: Suppose we have an upstairs valuation $$v$$ on $$L$$. Then it makes sense to restrict valuations to subfields, so \begin{align*} v\in \Sigma(L/B) \iff { \left.{{v}} \right|_{{K}} } \in \Sigma(K/A) .\end{align*} So the original valuation is $$B{\hbox{-}}$$regular iff the restricted valuation is $$A{\hbox{-}}$$regular.

$$\impliedby$$: Since $$A \subseteq B$$, being $$B{\hbox{-}}$$regular implies being $$A{\hbox{-}}$$regular.

$$\implies$$: Suppose $$A \subset R_v$$ and $$x\in B$$, and choose $$a_0, \cdots, a_{n-1} \in A$$ such that \begin{align*} p(x) \mathrel{\vcenter{:}}= x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0 .\end{align*} We can do this precisely because $$B$$ is integral over $$A$$. So we have an integral relation for $$x$$, and we want to show $$v(x) < 0$$ and derive some contradiction from the fact that $$v(a_i) \geq 0$$. Note that we aren’t grounded to the base field here, so this valuation may not be discrete and is rather some arbitrary Krull valuation.

If $$x\not \in R_v$$, then $$v(x) < 0$$, and we can thus write \begin{align*} v(x^n) < \min \left\{{ v(a_j x^j) {~\mathrel{\Big|}~}0\leq j \leq n-1 }\right\} \leq v(p(x)) .\end{align*} This follows because the first term is $$nv(x)$$, and so the next term can only be less negative since $$v(a_j) > 0$$. But this is a contradiction, since we know $$v(x^n) = v(- \sum_{j=0}^{n-1} a_j x^j)$$, and we’ve exhibited two elements that differ by a unit ($$u=-1$$) which have different valuations.

Next, let $$K/k$$ be a one variable function, we want to give a nice description of its places. We already described the places of a rational function field, and we know we can write the former function fields as finite degree extensions of the latter. Choosing a transcendental $$t\in K$$, to $$K/k(t)$$ is a finite extension, restricting evaluations gives a map \begin{align*} r: \Sigma(K/k) \to \Sigma(k(t)/ k) .\end{align*}

This is surjective with finite fibers, so it acts like a branched covering map.

This follows from NTI or NTII. The NTI method is taking an extension of Dedekind domains, taking a prime ideal downstairs, and pushing it forward to see how it factors upstairs. The NTII method is a field with a valuation and an extension of the field and you try to figure out how many ways the downstairs valuation can be extended. If the valuations are discrete, these are the same problem.

## 7.5 An Inequality on Degrees

Let $$K$$ be a field with $$v$$ a rank one valuation with valuation ring $$R$$. Let $$L/K$$ be a finite extension of degree $$n$$. Then the set of valuations on $$L$$ extending $$v$$ is finite and nonempty, say $$\left\{{w_1, \cdots, w_g}\right\}$$.

For $$1\leq i \leq g$$, define \begin{align*} e_i(L/K) &\mathrel{\vcenter{:}}={\left\lvert { w_j (L^{\times}) \over v(K^{\times}) } \right\rvert} && \text{ramification index} \\ f_i(L/K) &\mathrel{\vcenter{:}}=[R_{w_i}/\mathfrak{m}_{w_i} : R_v/ \mathfrak{m}_v] && \text{residual degree} ,\end{align*} so $$e_i, f_i \in {\mathbb{Z}}{>0}$$. Then

1. We have a useful inequality: \begin{align*} \sum_{i=1}^g e_i(L/K) f_i(L/K) \leq [L: K] = n .\end{align*}

2. If $$v$$ is discrete29 and the integral closure $$S$$ of $$R$$ in $$L$$ is finitely generated as an $$R{\hbox{-}}$$module, then this is an equality.

Note that a valuation can be extended in at least one way over any field extension, finite or not. For finite extensions, there’s a more precise statement involving completing and taking a tensor product, then identifying number of valuations with the size of some $$\operatorname{mSpec}$$ over a finite-dimensional algebra over the field.

NTII shows that $$e_i$$ is a finite number by looking at the exponent of the pushforward. Also note that we view $$\mathfrak{m}_{w_i}$$ as an ideal lying over $$\mathfrak{m}_v$$, and there is an inclusion of residue fields $$R_v/\mathfrak{m}_v \hookrightarrow R_{w_i} / \mathfrak{m}_{w_i}$$ which is in fact a finite degree field extension.

Part (a) already shows that $$r$$ is surjective with fibers of cardinality at most $$[L: K]$$, but we want equality. We claim is always holds when $$K/k$$ is a one variable function field and $$v\in \Sigma(K/k)$$. There are examples where the inequality is strict, however. In our situation, it’s not just an arbitrary extension, we have the aforementioned affine “grounding” phenomenon, and all of these DVRs are going to be localizations of affine Dedekind domains. This is the key fact: arbitrary extensions of Dedekind domains are nowhere near as nice as those where the bottom one is finitely generated over a field.

We have a discrete valuation $$v$$ on $$K$$, so let $$t$$ be a uniformizing element30 for $$v$$. Then the argument is that any such uniformizer $$t$$ is transcendental over $$k$$. We’ll do this by arguing $$t\not\in k$$ and then that $$t$$ is not algebraic over $$k$$ either.

Since we’re assuming $$v$$ is $$k{\hbox{-}}$$regular, $$t\in k \implies 1/t\in k$$ and so $$v(1/t) \geq 0$$, since every element in $$k$$ should have nonnegative valuation. But we’re supposed to have $$v(t) = 1$$ by definition of being a uniformizer, so $$t$$ can not be in $$k$$.

Suppose that $$t$$ is algebraic over $$k$$, then $$k(t)/k$$ is an integral extension, since we’re adjoining one algebraic element. By the previous proposition we have that $$v$$ is $$k(t){\hbox{-}}$$regular, since being regular is preserved by integral extensions. But now rerunning the argument in the previous paragraph shows that this is a contradiction: being $$k(t){\hbox{-}}$$regular would force $$v(1/t) \geq 0$$, but we’d still need $$v(1/t) = -1$$.

So $$t$$ is transcendental over $$k$$, and $$k[t]$$ is a polynomial ring.

Let

• $$A$$ be the integral closure of $$k[t]$$ in $$K$$, and
• $$B$$ be the integral closure of $$k[t]$$ in $$L$$.

Instead of a NTI square, we’ll have the following 3-step diagram: So $$A$$ is a Dedekind domain with $$\operatorname{ff}(A) = K$$, as is $$B$$ with $$\operatorname{ff}(B) = L$$, making both $$A$$ and $$B$$ finitely generated $$k[t]{\hbox{-}}$$modules. Why? This comes from the theorem of finiteness of integral closure when the downstairs domain is grounded to a field. Since $$k[t]$$ is finitely generated as a $$k{\hbox{-}}$$algebra, this finiteness applies, which tells us that $$A$$ finitely generated as a $$k[t]{\hbox{-}}$$module, as is $$B$$. But if $$B$$ is finitely generated as a $$k[t]{\hbox{-}}$$module and $$A\supseteq k[t]$$ is an even larger ring, then $$B$$ is finitely generated as an $$A{\hbox{-}}$$module (potentially with fewer generators).

Thus $$B$$ is a finitely generated $$A{\hbox{-}}$$module, and $$v$$ is $$k[t]{\hbox{-}}$$regular since $$t$$ was a uniformizing element, making $$v$$ regular on both $$k$$ and $$t$$ and thus $$k[t]$$. Then $$v$$ is also $$A{\hbox{-}}$$regular by the proposition, and thus $$v = v_{\mathfrak{p}}$$ for some $$\mathfrak{p}\in \operatorname{mSpec}(A)$$ coming from our classification of $$A{\hbox{-}}$$regular valuations on a Dedekind domain.

So the valuation on $$K$$ is just the $$\mathfrak{p}{\hbox{-}}$$adic valuation on this Dedekind domain. This means there is an equality of valuation rings $$R = A_{\mathfrak{p}}$$31, the valuation ring of the Dedekind domain. So we now consider $$S$$, the integral closure of $$R$$ in $$L$$. This is a NTI situation, but the downstairs Dedekind domain is a DVR, so it’s local downstairs. We thus have compatibility between integral closure and localization in the form of $$S = B_{\mathfrak{p}} = B \otimes_A A_{\mathfrak{p}}$$. This comes from taking the whole integral closure $$B$$, and only looking at the primes lying over $$\mathfrak{p}$$. Base change preserves finite generation, and we know that $$B$$ was finitely generated as an $$A{\hbox{-}}$$module, so $$S$$ is finitely generated as an $$A_{\mathfrak{p}}{\hbox{-}}$$module and equality holds.

If $$A_{\mathfrak{p}}$$ was a complete DVR, as opposed to just some localization of an affine domain, $$B$$ will be a semilocal Dedekind domain and thus a PID, and again the number of primes it has will be the number of primes in the original Dedekind domain lying over the fixed prime $$\mathfrak{p}$$.

We’re not really using valuation theory here, and this could have been phrased purely in NTI language. But even then, the degree inequality for extensions of Dedekind domains needs finite generated of the Dedekind domain as a module over the bottom Dedekind domain to ensure equality. You’d need a suitably algebraic text that considers not necessarily separable $$L/K$$, and you really do want finite generation of $$B$$ over $$A$$ to make this work. See Dino Lorenzini’s textbook!

Let $$K/k$$ be a one variable function field, and show that the cardinality of the set of points is given by \begin{align*} {\left\lvert {\Sigma (K/k)} \right\rvert} = {\left\lvert {\left\{{ \text{monic irreducible polynomials } p \in k[t] }\right\}} \right\rvert} = \max( {\left\lvert {k} \right\rvert}, \aleph_0 ) .\end{align*}

If you know that $$r$$ is surjective with finite fibers, where the image is infinite (which it is here), the domain should be infinite of the same cardinality by an easy set-theoretic exercise. Note that using Möbius inversion, over a finite field there is at least one irreducible polynomial of every degree, and finitely many of a fixed degree. So the cardinality is $$\aleph_0$$ when $$k$$ is a finite field. If we took a one variable function field over $${\mathbb{C}}$$, we would get the cardinality of the continuum. In this case, $$\Sigma(K/k)$$ really is the set of points on some compact Riemann surface, although the Zariski topology will be too coarse to coincide with the induced Euclidean topology.

Note that affine Dedekind domains are important for us because every finitely generated field extension of $$k$$ are precisely the fraction fields of affine domains over $$k$$, where the transcendence degree of the function field equals the Krull dimension of the affine domain. We’re especially interested in affine domains of dimension 1 over $$k$$. We established something particularly important in this proof:

## 7.6 Affine Grounding and Residue Fields

Let $$K/k$$ be a one variable function field and $$v\in \Sigma(K/k)$$ be a place on that function field. Then there exists an affine Dedekind domain $$A$$ with $$\operatorname{ff}(A) = K$$ and a maximal ideal $$\mathfrak{p}\in \operatorname{mSpec}(A)$$ such that $$R_v = A_{\mathfrak{p}}$$.

Thus we should think of the set of places as the $$\operatorname{mSpec}$$ of finitely many affine Dedekind domains glued together. For each point (place), the basic open set around that point is the affine Dedekind domain.

For $$v \in \Sigma(K/k)$$, define the residue field of the local ring $$R_v$$ as $$k(v) \mathrel{\vcenter{:}}= R_v / \mathfrak{m}_v$$. Then $$k(v)/ k$$ is a finite degree extension.

If $$R$$ is a domain with maximal ideal $$\mathfrak{p}$$, then the quotient map factors through the localization, giving $$R/\mathfrak{p} = R_{\mathfrak{p}} / \mathfrak{p} R_{\mathfrak{p}}$$:32 So by affine grounding, $$k(v)$$ is also $$A/\mathfrak{p}$$ where $$A$$ is an affine Dedekind domain and $$\mathfrak{p}\in \operatorname{mSpec}(A)$$. This is Zariski’s lemma33 : we showed that $$k(v) \cong A/\mathfrak{p}$$, where $$A$$ is a finitely generated algebra and thus so are its quotients. Thus $$k(v)$$ is not just finitely generated as a field extension, but also as a $$k{\hbox{-}}$$algebra, making $$k(v)/k$$ a finite extension.

The degree of $$v\in \Sigma(K/k)$$ is $$[k(v) : k] \in {\mathbb{Z}}^{\geq 0}$$.

We are especially interested in degree 1 places, i.e. those for which the residue field is equal to $$k$$ itself, so we denote these by $$\Sigma_1(K/k)$$. In any other course, we’d call this $$C(k)$$, the rational points on the associated curve.

Let $$f\in k[x, y]$$ be irreducible, so that $$A \mathrel{\vcenter{:}}= k[x, y] / \left\langle{f}\right\rangle$$ is a 1-dimensional affine domain34. As above, the resude fields of maximal ideals are finite extensions of $$k$$. Show that there is a correspondence \begin{align*} \left\{{\substack{\text{ Maximal ideals } \\ \mathfrak{p} \in \operatorname{mSpec}(A)}}\right\} &\iff \left\{{\substack{(x, y) \in k\times k {~\mathrel{\Big|}~}f(x, y) = 0}}\right\} .\end{align*}

Note that the polynomial above may not define a smooth geometry, there may instead be singular points:

These singular points are what stops $$A$$ from being integrally closed, which is literally true when $$k$$ is a perfect field.

Whereas $$\Sigma(K/k)$$ is always infinite, $$\Sigma_1(K/k)$$ may be finite or even empty. When $$k = {\mathbb{Q}}$$, it may in fact be empty “most of the time” When $$k = {\mathbb{Q}}$$, it may in fact be empty “most of the time.”

For all $$v\in \Sigma(K/k)$$, the degree of the point $$\deg(v)$$ will be divisible by $$[\kappa(K) : k]$$. Thus if $$\kappa(L) \supsetneq k$$, then $$\Sigma_1(K/k) = \emptyset$$.35

Note that before we were writing the residue field as an extension of $$k$$, and it’s worth checking that the constant subfield embeds as a subfield of the residue field as well.

There is a tie to CM points on modular curves: if you have a function field over $${\mathbb{Q}}$$ which is not regular due to some proper algebraic subextension, the residue fields of all of the points on the curve will contain the algebraic closure of the field of definition. Pete had some $${\mathbb{Q}}(X_n)$$ function field, whose constant subfield was $${\mathbb{Q}}(\zeta_n)$$ (adjoining the $$n$$th roots of unity), and none of these modular curves over $${\mathbb{Q}}$$ have closed points except when the residue fields contain $${\mathbb{Q}}(\zeta_n)$$.

This is a way for there to not be points on the curve, so $$\Sigma_1(K/k)$$ is empty, but it’s not the deepest reason – this is a cheap trick to produce “pointless” function fields. It can fail to have degree 1 places in many different ways!

Show that for for a one variable function field $$K/k$$ TFAE:

1. Every $$v\in \Sigma(K/k)$$ has degree 1,

2. $$k$$ is algebraically closed.

One half is easy, since by definition the degree of the residue field is the degree of some finite extension of the base field, but if $$k$$ is algebraically closed, the degree of any finite extension is one.

For a field $$k$$, set $${\mathbb{P}}^1(k) \mathrel{\vcenter{:}}={\mathbb{P}}(k^2)$$, the projectivization of $$k\times k$$, i.e. the lines through the origin in $${\mathbb{A}}^2/k$$. By taking slopes of lines, $${\mathbb{P}}^1(k) = k {\coprod}\left\{{\infty}\right\}$$.

Show that $$\Sigma_1(k(t)/k) = {\mathbb{P}}^1(k)$$, and deduce that \begin{align*} \Sigma(k(t)/k) = {\mathbb{P}}^1(k) \iff k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu .\end{align*}

Next up we’ll talk about how the set of places is built from affine Dedekind domains. After this, we’ll be ready for chapter 2: divisors and Riemann-Roch.

# 8 Lecture 6: Affine Domains and Places $$\Sigma(K/k)$$

The aim of this lecture is to explain the difference (including some technicalities) between $$\Sigma(K/k)$$ and affine Dedekind domains $$R$$ such that $$K = \operatorname{ff}(R)$$.

Recall that

• An affine domain over a field $$k$$ is a domain that is finitely generated as a $$k{\hbox{-}}$$algebra,36

• An affine Dedekind domain is an affine domain that is also a Dedekind domain, so it is integrally closed and of Krull dimension 1,

• An affine $$k{\hbox{-}}$$order is a one-dimensional affine domain.37

If $$f \in k[x, y]$$ is irreducible, then $$k[x, y]/\left\langle{f}\right\rangle$$ is an affine $$k{\hbox{-}}$$order. It is an affine Dedekind domain if $$f$$ is nonsingular over $$k$$, i.e. for all $$a, b\in \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu$$ such that $$f(a, b) = 0$$, the usual partial derivatives in the sense of Calculus $${\frac{\partial f}{\partial x}\,}$$ and $${\frac{\partial f}{\partial y}\,}$$ do not simultaneously vanish at $$(a, b)$$. This is a sufficient condition, although it’s not far from being necessary as well.

Let $$A/k$$ be an affine Dedekind domain such that $$\operatorname{ff}(A) = K$$. Then $$\operatorname{mSpec}(A) = \Sigma(K/A) \hookrightarrow\Sigma(K/k)$$. This follows because $$\Sigma(K/A)$$ are the valuations that are not just regular on $$k$$, but also on $$A$$, (i.e. $$A{\hbox{-}}$$regular valuations) so the valuation ring contains the entirety of $$A$$. It’s thus natural to ask what its complement is, i.e. those valuations which are not regular on $$A$$ and give its elements negative valuation. So define \begin{align*} \Sigma(A, \infty) \mathrel{\vcenter{:}}=\Sigma(K/k) \setminus\Sigma(K/A) ,\end{align*} the set of places at infinity with respect to $$A$$.

$$\Sigma(k[t], \infty) = \left\{{v_ \infty}\right\}$$, which is the infinite place, so the terminology at least matches up!

For any affine Dedekind domain $$A$$, $$\Sigma(A, \infty)$$ is finite and nonempty.

This is striking! This says that one affine Dedekind domain is giving almost all of this infinite set of places, but never all of it.

By Noether Normalization38

there exists a $$t\in A$$ that that $$A$$ is a finitely generated (and thus integral) $$k[t]{\hbox{-}}$$module, and $$A$$ is the integral closure of $$k[t]$$ in $$K$$. Why must this be the integral closure? Any ring finitely generated over a subring will be an integral extension, and $$A$$ is a Dedekind domain and thus integrally closed. So let \begin{align*} r: \Sigma(K/k) \to \Sigma(k(t)/k) \end{align*} denote the restriction map; then by the regularity property we established in , we have \begin{align*} \Sigma(K/A) = r^{-1}\qty{\Sigma(k(t)/k[t])} .\end{align*} Why? A valuation upstairs in the NTI square is regular with respect to the integral extension upstairs iff it’s regular with respect to the ring it is the integral extension of. So regularity is preserved both ways by integral extensions. This means you can check regularity either upstairs or downstairs, allowing us to identify the above preimage.

This means that the places where are not $$A{\hbox{-}}$$regular upstairs are precisely those which are not $$k[t]{\hbox{-}}$$regular downstairs, and so we have \begin{align*} \Sigma(A, \infty) = r^{-1}\qty{\Sigma(k[t], \infty)} = r^{-1}(v_ \infty) ,\end{align*} since we now there is exactly one such non-regular valuation. But we showed that $$r$$ was surjective with finite nonempty fibers, so we’re done since our set is one of the fibers.

Thus is $$K/k$$ is a one variable function field and $$A$$ is an affine Dedekind domain with fraction field $$K$$, then $$\Sigma(K/k) = \operatorname{mSpec}(A) {\coprod}S$$ where $$S$$ is finite and nonempty. Earlier we saw by affine grounding that for each $$v\in \Sigma(K/k)$$ there exists an affine Dedekind domain $$A$$ with $$v\in \Sigma(K/A)$$, and thus $$\Sigma(K/k)$$ admits a finite covering by $$\operatorname{mSpec}$$ of affine Dedekind domains. The picture of what’s happening is that we have $$\Sigma(K/k)$$ which is quasicompact with respect to the Zariski topology, which contains many $$\operatorname{mSpec}$$, at least one of which contains $$v$$. Note that these $$\operatorname{mSpec}(A_j)$$ for affine Dedekind domains $$A_j$$ is literally an open cover in this topology. But the open sets are so large that they all have finite complement. However, this means that instead of just an arbitrary open covering, one can choose a finite open covering: one $$\operatorname{mSpec}(A_j)$$ will cover all but finitely many, and we can always find at least one $$\operatorname{mSpec}(A_{j'})$$ covering all of the remaining points.